Partial Fractions

In Calculus, there are several procedures that are much easier if we can take a rather large fraction and break it up into pieces. The procedure that can decompose larger fractions is called Partial Fraction Decompostition. We will proceed as if we are working backwards through an addition of fractions with LCD.

EXAMPLE 1: For our first example we will work an LCD problem frontwards and backwards. Use an LCD to complete the following addition.

21

78

21

10533

2

2

1

5

2

3

1

11

5

2

3

xx

x

xx

xx

x

x

xxx

xxx

1

5

2

3

xx

The LCD is (x + 2)(x – 1). We now convert each fraction to LCD status.On the next slide we will work this problem backwards

2

782

xx

xFind the partial fraction decomposition for:

As we saw in the previous slide the denominator factors as (x + 2)(x – 1). We want to find numbers A and B so that:

122

782

x

B

x

A

xx

x

The bad news is that we have to do this without peeking at the previous slide to see the answer. What do you think will be our first move?

So we multiply both sides of the equation by (x + 2)(x – 1).

Now we expand and compare the left side to the right side.

If the left side and the right side are going to be equal then:

A+B has to be 8 and

-A+2B has to be 7.

BAxBAx

BBxAAxx

xBxAx

xxx

Bxx

x

A

xx

xxx

xxx

B

x

A

xx

xxx

278

278

2178

121

12212

7812

12122

7812

2

A + B = 8

-A + 2B = 7

This gives us two equations in two unknowns. We can add the two equations and finish it off with back substitution.

3B = 15

B = 5

If B = 5 and A + B = 8 then A = 3.

Cool!! But what does this mean?

Remember that our original mission was to break a big fraction into a couple of pieces. In particular to find A and B so that:

122

782

x

B

x

A

xx

x

We now know that A = 3 and B = 5 which means that

1

5

2

3

2

782

xxxx

x And that is partial fraction decomposition!

8

x 1 x 3

A B 8

x 1 x 3 x 1 x 3

2

8

x 2x 3

A B x3 80x x 1

A

8

B

B

0

3A

A 2, B 2

2 2

x 1 x 3

2 2

x 1 x 3

2 x 3 2 x 1

x 1 x 3 x 3 x 1

2 x 3 2 x 1

x 1 x 3

2

8

x 2x 3

Find the partial fraction decomposition of

5x 7

x 2 x 1

A B 5x 7

x 2 x 1 x 2 x 1

A B x1 75x x 2

A 2B

A B 5

7

A 3,B 2

2

5x 7Find the partial fraction decomposition of

x 3x 2

3 2

x 2 x 1

2

8x 4

x 2x 3

2

8x 4 A B

x 2x 3 x 3 x 1

8 A x4 3Bx x 1

A 3B

A B 8

4

A 7, B 1

7 1

x 3 x 1

Using the method of partial fractions to decompose2

5x 1

x 1

2

A B 5x 1

x 1 x 1 x 1

A B x1 15x x 1

A B 5

A B 1

A 3, B 2

3 2

x 1 x 1

2

1Find the partial fraction decomposition of

x x

2

A B 1

x x 1 x x

x x1 1A B

A

A B

1

0

1 1

x x 1

1

x x 1

A 1, B 1

EXAMPLE 2: Find the partial fraction decomposition for

1243

101123

2

xxx

xx

First we will see if the denominator factors. (If it doesn’t we are doomed.)

The denominator has four terms so we will try to factor by grouping.

43

34312432

223

xx

xxxxxx

Since the denominator is factorable we can pursue the decomposition.

3434

1011

1243

101122

2

23

2

x

C

x

BAx

xx

xx

xxx

xx

Because one of the factors in the denominator is quadratic, it is quite possible that its numerator could have an x term and a constant term—thus the use of Ax + B in the numerator.

3434

101122

2

x

C

x

BAx

xx

xx As in the first example, we multiply both sides of this equation by the LCD.

)43()3(1011

4331011

343

344

1011

343434

101134

22

222

222

2

222

22

CBxBAxCAxx

CCxBBxAxAxxx

xxx

Cxx

x

BAxxx

xxx

C

x

BAx

xx

xxxx

If the two sides of this equation are indeed equal, then the corresponding coefficients will have to agree:

-1 = A + C

11 = 3A + B

-10 = 3B + 4C

On the next slide, we solve this system. We will start by combining the first two equations to eliminate A.

-1 = A + C

11 = 3A + B

-10 = 3B + 4C

3 = -3A - 3C

11 = 3A + B

Multiply both sides by -3

Add these two equations to eliminate A.

14 = B – 3C

Multiply both sides of this equation by –3.

Add this equation to eliminate B.

-42 = -3B + 9C-10 = 3B + 4C

-52 = 13C

-4 = C

We now have two equations in B and C. Compare the B coefficients.

We can finish by back substitution.

-1 = A + C -1 = A - 4 A = 3

-10 = 3B + 4C -10 = 3B + 4(-4) -10 + 16 = 3B 2 = B

We have now discovered that A = 3, B = 2 and C = -4.

OK, but I forgot what this means.

Fair enough. We began with the idea that we could break the following fraction up into smaller pieces (partial fraction decomposition).

3443

1011

1243

1011

22

2

23

2

x

C

x

BAx

xx

xx

xxx

xx

Substitute for A, B and C and we are done.

3

4

4

232

xx

x

2

2

x 11x 18Find the partial fraction decomposition of

x x 3x 3

2

2 2

A Bx C x 11x 18

x x 3x 3 x x 3x 3

2 2A x 3x 3 Bx C x x 11x 18

A B 1

3A C 11

3A 18

2

6 7x 7

x x 3x 3

A 6, B 7, C 7

2

2

3x x 4Find the partial fraction decomposition of

x x 1

2

2 2

A Bx C 3x x 4

x x 1 x x 1

2 2A x 1 x Bx C 3x x 4

A B 3

C 1 A 4,B 1, C 1

A 4

2

4 x 1

x x 1

2

4Find the partial fraction decomposition of

x x 1

2 2

A Bx C 4

x x 1 x x 1

2A x 1 Bx C x 4

A B 0

C 0

A 4

2

4 4x

x x 1

A 4, B 4, C 0

EXAMPLE 3: For our next example, we are going to consider what happens when one of the factors in the denominator is raised to a power. Consider the following for partial fraction decomposition:

22

2

2

23

2

3

724813

96

724813

96

724813

xx

xx

xxx

xx

xx

xx

There are two setups that we could use to begin:

Setup A proceeds along the same lines as the previous example.

22

2

)3(3

724813

x

CBx

x

A

xx

xx

Setup B considers that the second fraction could have come from two pieces.

22

2

)3(33

724813

x

C

x

B

x

A

xx

xx

AxCBAxBAxx

CxBxBxAAxAxxx

CxBxBxxxAxx

CxxBxxAxx

xxx

Cxx

x

Bxx

x

Axx

xxx

C

x

B

x

A

xx

xxxx

936724813

396724813

396724813

33724813

33

33

3724813

3)3(33

7248133

22

222

222

22

2

2

222

2

22

22

Since we have already done an example with Setup A, this example will proceed with Setup B. Step 1 will be to multiply both sides by the LCD and simplify.

Expand.

Group like terms and factor.

We now compare the coefficients of the two sides.

AxCBAxBAxx 936724813 22 The last line of the previous slide left us here.

If we compare the coefficients on each side, we have:

A + B = 13

6A + 3B + C = 48

9A = 72From the third equation A = 8. Substituting into the first equation:

A + B = 13 so 8 + B = 13 and B = 5.

Substituting back into the second equation:

6A + 3B + C = 48 so 6(8) + 3(5) + C = 48

48 + 15 + C = 48 63 + C = 48 and C = -15

22

2

)3(33

724813

x

C

x

B

x

A

xx

xx

To refresh your memory, we were looking for values of of A, B and C that would satisfy the partial fraction decomposition below and we did find that A= 8, B=5 and C=-15.

So…..

22

2

)3(

15

3

58

3

724813

xxxxx

xx

2

x 4Find the partial fraction decomposition of

x 1

2 2

A B x 4

x 1 x 1 x 1

A B1 4x x

A B

1

4

A

2

1 3

x 1 x 1

A 1, B 3

EXAMPLE 4: Find the partial fraction decomposition for

82

515422

23

xx

xxx

Since the order of the numerator is larger than the order of the denominator, the first step is division.

5

1642

82

52

5154282

23

2

232

x

xxx

xx

xx

xxxxx

By long division we have discovered that:

82

52

82

5154222

23

xx

xx

xx

xxx

We will now do partial fraction decomposition on the remainder.

BAxBAx

BBxAAxx

xBxAx

xxx

Bxx

x

Ax

xxx

B

x

A

xx

xxx

x

B

x

A

xx

x

xx

x

425

425

425

242

244

5

242424

524

2424

5

82

52

Multiply both sides by the LCD.

Distribute

Group like terms

Compare coefficients

From the previous slide we have that:

BAxBAx 425

If these two sides are equal then:

1 = A + B and 5 = 2A – 4B

To eliminate A multiply both sides of the first equation by –2 and add.

2A – 4B = 5

-2A – 2B = -2

-6B = 3 so B = -1/2

If A + B = 1 and B = -1/2 then

A –1/2 = 2/2 and A = 3/2

22

1

42

32

2

2/1

4

2/32

242

82

52

82

5154222

23

xxx

xxx

x

B

x

Ax

xx

xx

xx

xxx

In summary then: